The Function approximate the velocity (in feet per second) of an object dropped from a height of x feet right before it hits the ground on Earth and on Mars.
Earth: E(X) = 8V
Mars: M(x) = 0.6(E(x))
What is the velocity of an object dropped from a height og 25 feet right before it hits the ground on
Mars?

The answer to the question is

It take for the golf ball hit on Mars to reach the ground 8.284 s longer.

Step-by-step explanation:

To Solve the question, we note that the equation of motion is given to us as

h = -1.9·t² + 26·t

Therefore the height

We compare the above equation with the equation of of motion

S = ut - 0.5×g×t² where u = initial velocity = 26 m/s and  - 0.5×g = -1.9

Therefore the acceleration due to gravity om mars = -1.9/-0.5 = 3.8 m/s²

We now have the time taken to maximum height given by

v = u - g·t where v = 0 at maximum height

We therefore have u = g·t → 26 = 3.8×t or t = 26/3.8 = 6.84 seconds

The time taken for ball to hit the ground is 2 × the time to reach maximum height = 2 × 6.84 = 13.684 s

The time longer it takes for the ball on Mars to complete the path up to maximum height and down again is 13.684 s - 5.4 s = 8.284 s.

the answer is 1030 it's very simple math.